Balance the equation in aqueous basic solution:
As2S3(s) + H2O2(aq) → AsO43-(aq) + SO42-(aq)
I normally understand how to balance redox equations, but this one confuses me because I would normally think that the As2S3 is being reduced to AsO4^3-, but since the As2S3 has a "S", do i also include the SO4^2-? but even when i do that, and try to balance the electrons for the half rxn, it seems to change into oxidation... I'm really confused!
Here’s a hint. The half reactions that must be balanced are As₂S₃ —> AsO₄³⁻ + SO₄²⁻ (oxidation) and H₂O₂ —> ? (reduction). You can determine which ions to group by examining oxidation numbers. Note that since H₂O₂ consists of only hydrogen and oxygen, you can instantly balance the half reaction by adding water and H⁺/OH⁻ ions as you would for any half reaction once the non-hydrogen/non-oxygen species have been balanced by inspection. This explains why there is no obvious product in the half reaction (the product is water which is added in only when balancing).
I usually attack this type this way. Yes, both As and S are changing, As from +6(total of 2 atoms As) to +10(total of 2 atoms As). S is changing from -6 (total of 3 S atoms to +18(total of 3 atoms S). Most of the time it works out if one simply combines the As and S into one kind of reaction (add electrons algebraically). What has been omitted is the reaction of the H2O2. You'll see what to do when that is included.
Ah, balancing redox equations can be a tricky business, but fear not, I am here to help you through the confusion, even if the subject matter is a little "sulfurous." So, let's break it down step by step:
First, let's assign oxidation states to the elements involved. We have As in As2S3, which has an oxidation state of 0, and S in As2S3, which has an oxidation state of -2.
Next, we examine the product AsO43-. Since As is a central atom in this polyatomic ion, it must have an oxidation state of +5. This means that As is being oxidized from an oxidation state of 0 to +5.
Now, let's look at S in SO42-. Since oxygen is more electronegative and takes priority for assigning oxidation states, we know that the oxidation state of S in SO42- is +6. This means that S is being oxidized from an oxidation state of -2 to +6.
So, to balance the equation, we need to split it into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Let's start with the oxidation half-reaction:
As2S3(s) → AsO43-(aq)
Here we have the arsenic in As2S3 being oxidized, so let's balance it by adding electrons:
As2S3(s) → 2AsO43-(aq) + 6e-
Now let's move on to the reduction half-reaction:
H2O2(aq) → SO42-(aq)
The sulfur in As2S3 is being oxidized, so let's balance it by adding water:
H2O2(aq) + 2H2O(l) → SO42-(aq)
Now, to balance the oxygen atoms, we need to add hydroxide ions (OH-) to both sides:
H2O2(aq) + 2H2O(l) + 4OH-(aq) → SO42-(aq) + 4OH-(aq)
Finally, we can combine the two half-reactions and cancel out the spectator ions:
As2S3(s) + 2H2O2(aq) + 2H2O(l) + 4OH-(aq) → 2AsO43-(aq) + SO42-(aq) + 4OH-(aq)
And there you have it, the balanced equation in aqueous basic solution! Just remember, when dealing with redox reactions, keep a positive attitude, use your chemical "wits," and you'll "balance" through any confusion!
Balancing redox equations can sometimes be challenging, but with a systematic approach, we can solve it step by step.
1. Start by separating the equation into two half-reactions, oxidation and reduction:
a. Oxidation half-reaction: As2S3(s) → AsO43-(aq)
b. Reduction half-reaction: H2O2(aq) → SO42-(aq)
2. Let's begin by balancing the atoms that are not involved in redox reactions. In this case, we focus on the elements other than As, S, O, and H.
- As2S3(s) does not contain any atoms to balance.
- H2O2(aq) contains 2 hydrogen (H) atoms on the left side.
3. Balance the atoms undergoing redox reactions.
- As: To balance As atoms, we need to multiply the AsO43- by 2 in the oxidation half-reaction.
Oxidation half-reaction: As2S3(s) → 2AsO43-(aq)
- S: One As2S3 is on the left side, while three sulfurs (S) are on the right side. To balance these, multiply H2O2(aq) by 3 in the reduction half-reaction.
Reduction half-reaction: 3H2O2(aq) → 3SO42-(aq)
4. Now, let's balance the charge by adding electrons (e-) to the appropriate side of each half-reaction.
- Balance the oxidation half-reaction:
As2S3(s) → 2AsO43-(aq) + 6e-
- Balance the reduction half-reaction:
3H2O2(aq) + 6e- → 3SO42-(aq)
5. To balance the electrons, equalize the number of electrons between the two half-reactions by multiplying one or both of the half-reactions.
- Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons:
3(As2S3(s) → 2AsO43-(aq) + 6e-)
2(3H2O2(aq) + 6e- → 3SO42-(aq))
Simplified equation:
3As2S3(s) + 6H2O2(aq) → 2AsO43-(aq) + 6SO42-(aq)
6. Finally, balance the remaining elements, which are hydrogen (H) and oxygen (O).
- Hydrogen: There are 12 hydrogen (H) atoms on the right side and 12 hydrogen (H) atoms on the left side. They are already balanced.
- Oxygen: There are ten oxygen (O) atoms on the right side and six oxygen (O) atoms on the left side. To balance the oxygen, add four water (H2O) molecules to the left side.
3As2S3(s) + 6H2O2(aq) → 2AsO43-(aq) + 6SO42-(aq) + 4H2O(l)
We have now balanced the equation in aqueous basic solution:
3As2S3(s) + 6H2O2(aq) → 2AsO43-(aq) + 6SO42-(aq) + 4H2O(l)